Hi, is there any other routine for factor analysis in R then factanal? Basically I'am interested in another extraction method then the maximum likelihood method and looking for unweighted least squares. Thanks in advance Sigbert Klinke
I wrote some rough functions for principal factor, principal-components factor, and iterated principal factor analysis. I think they are workable, the same results as stata can be retained. In addition, functions for gls and uls factor analysis is in progress, which is based on the algorithms of SPSS. I get the same results by the gls factor analysis, and quite similiar result by the uls factor analysis. 2007/6/1, Sigbert Klinke <sigbert at wiwi.hu-berlin.de>:> Hi, > > is there any other routine for factor analysis in R then factanal? > Basically I'am interested in another extraction method then the maximum > likelihood method and looking for unweighted least squares. > > Thanks in advance > > Sigbert Klinke > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >-- Ronggui Huang Department of Sociology Fudan University, Shanghai, China
Hi there, i?ve trouble understanding the factanal output of R. i am running a a FA on a dataset with 10 variables. i plotted eigenvalues to finde out how many factors to try. i think the "elbow" is @ 3 factors. here are my eigenvalues: 2.6372766 1.5137754 1.0188919 0.8986154 0.8327583 0.7187473 0.6932792 0.5807489 0.5709594 0.5349477 (of the correlation matrix) i guess this is basically what screeplot does as well. and here?s my problem: unfortunately the cumulative variance @ 3 factors is only .357 there are no crossloadings and the interpretation of the factors and their loadings definetely make sense so far. Can i use this factor analysis somehow despite the poor cumulative variance of the first three factors ? changing the rotation didnt help much. The test of the hypothesis says the following: Test of the hypothesis that 3 factors are sufficient. The chi square statistic is 46.58 on 18 degrees of freedom. The p-value is 0.000244 does this mean the Hnull is that 3 factors are sufficient and i cant recject ? 4 factors say: Test of the hypothesis that 4 factors are sufficient. The chi square statistic is 10.82 on 11 degrees of freedom. The p-value is 0.458 Unfortunately ?factanal does not tell me what the Hnull is in this case ? Thx a lot in advance for some advice matthias